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3 years ago

What are some of the uses of copper?

Chemistry

2 answers:

rusak2 [61]3 years ago

5 0

Body jewellery.
Belt buckles.
Cufflinks.
Bracelets.
Rings.
Necklaces.

Dmitrij [34]3 years ago

3 0

Answer:

Explanation:

Most copper is used in electrical equipment such as wiring and motors. This is because it conducts both heat and electricity very well, and can be drawn into wires. It also has uses in construction (for example roofing and plumbing), and industrial machinery (such as heat exchangers).

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What is the reason behind that the melting and boiling point of iron is more than that of sodium?

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Answer:

As metals are giant lattice structures, the number of electrostatic forces to be broken is extremely large, and so metals have high melting and boiling points. This means that the melting point and boiling point of metals are more similar to those for ionic compounds than for covalent substances.

Explanation:

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3 years ago

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Calcium hydroxide, Ca(OH)2, is an ionic compound with a solubility product constant, Ksp, of 6.5×10–6. Calculate the solubility

kari74 [83]

Answer: The solubility of this compound in pure water is 0.012 M

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of the is given as:

$Ca(OH)_2\rightarrow Ca^{2+}+2OH^-$

By stoichiometry of the reaction:

1 mole of $Ca(OH)_2$ gives 1 mole of $Ca^{2+}$ and 2 mole of $OH^-$

When the solubility of $Ca(OH)_2$ is S moles/liter, then the solubility of $Ca^{2+}$ will be S moles\liter and solubility of $OH^-$ will be 2S moles/liter.

$K_{sp}=[Ca^{2+}][OH^{-}]^2$

$6.5\times 10^{-6}=[S][2S]^2$

$S=0.012M$

Thus solubility of this compound in pure water is 0.012 M

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3 years ago

A) A 100-watt lightbulb radiates energy at a rate of 100 J/s. (The watt, a unit of power, or energy over time, is defined as 1 J

Ipatiy [6.2K]

Answer:

The answers are as given in the attachment

Explanation:

The application of the de brogile equation was used and appropriate substitution were made as shown in the attachment

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3 years ago

Suppose 11.4g of ammonium chloride is dissolved in 250 ml of a 0.3M aqueous solution of potassium carbonate. Calculate the final

Sindrei [870]

Answer:

The molarity of the final ammonium cation is 0.252M

Explanation:

Step 1: Data given

Mass of ammonium chloride (NH4Cl) = 11.4 grams

Volume of 0.3 M aqueous solution of potassium carbonate (K2CO3) = 250 mL = 0.250L

Step 2: The balanced equation

2NH4Cl + K2CO3 → 2KCl + (NH4)2CO3

Step 3: Calculate moles of (NH4)Cl

moles (NH4)Cl = 11.4 grams /53.49 g/mol

Moles (NH4)Cl = 0.213 moles

Step 4: Calculate moles of K2CO3

Moles K2CO3 = Molarity * Volume

Moles K2CO3 = 0.3M * 0.250 L = 0.075 moles

Step 5: Calculate moles (NH4)Cl at the equilibrium

For 2 moles (NH4)Cl consumed, we need 1 mole of K2CO3 to produce 2 KCl and 1 mole of (NH4)2CO3

(NH4)2CO3l will dissolve in 2NH4+ + CO32-

Moles (NH4)2Cl = 0.213 moles - 2*0.075 = 0.063 moles

Moles NH4+ = moles (NH4)Cl = 0.063 moles

Step 6: Calculate Molarity of NH4+

Molarity = Moles / volume

Molarity of NH4+ = 0.063 moles / 0.250 L

Molarity of NH4+ = 0.252 M

The molarity of the final ammonium cation is 0.252M

5 0

3 years ago

What are some of the uses of copper?​

What are some of the uses of copper?